These two equations concern a lab on the measurement of E/M by the Brainbridge Method.
I need to derive the absolute error for two different equations,
the first one is: e/m = 2V/B2R2
treat (e/m) as a SINGLE VARIABLE when doing the derivation.
The second is B=8μ0N(I-Io) / √125a
whre μo=4π*10-7 T*m/A
Io=zero referenc current
You may want to EXPLICITLY substitute in a new variable, (Inet), in place of (I – Io) [ie, let Inet = I – Io]
madherchod
To derive the absolute error for the first equation, where e/m = 2V/B^2R^2, treating e/m as a single variable, we can use the rules of error propagation.
Let's denote the absolute errors of the variables as Δe, Δm, ΔV, ΔB, ΔR.
The equation can be written as:
e/m = 2V / (B^2 * R^2)
To find the absolute error, we differentiate both sides of the equation with respect to each variable, while treating the others as constants.
Differentiating with respect to e:
∂(e/m) / ∂e = 1/m
Differentiating with respect to m:
∂(e/m) / ∂m = -e/m^2
Differentiating with respect to V:
∂(e/m) / ∂V = 2 / (B^2 * R^2)
Differentiating with respect to B:
∂(e/m) / ∂B = -4V / (B^3 * R^2)
Differentiating with respect to R:
∂(e/m) / ∂R = -4V / (B^2 * R^3)
Now, for each variable, we can multiply the absolute error by the partial derivative to get the contribution to the absolute error:
Δe = Δe/m * (∂(e/m) / ∂e)
= Δe/m * (1/m)
Δm = Δe/m * (∂(e/m) / ∂m)
= Δe/m * (-e/m^2)
ΔV = Δe/m * (∂(e/m) / ∂V)
= Δe/m * (2 / (B^2 * R^2))
ΔB = Δe/m * (∂(e/m) / ∂B)
= Δe/m * (-4V / (B^3 * R^2))
ΔR = Δe/m * (∂(e/m) / ∂R)
= Δe/m * (-4V / (B^2 * R^3))
These equations will give you the absolute errors for each variable in the equation e/m = 2V/B^2R^2, treating e/m as a single variable.
For the second equation, B = (8μ0N(I - Io)) / √(125a), where μ0 = 4π*10^(-7) T*m/A and Io is the zero-reference current.
You can substitute a new variable, Inet, in place of (I - Io), so let Inet = I - Io.
The equation then becomes: B = (8μ0NInet) / √(125a)
To derive the absolute error for B, we can use the same method as before.
Let ΔB, ΔInet, ΔN, and Δa be the absolute errors for each variable.
Differentiating the equation B = (8μ0NInet) / √(125a) with respect to each variable, while treating the others as constants:
∂B / ∂B = 1
∂B / ∂Inet = 8μ0N / √(125a)
∂B / ∂N = (8μ0Inet) / √(125a)
∂B / ∂a = (-4μ0NInet) / (2√(125a^3))
Using the same approach as before, we can calculate the absolute errors for each variable:
ΔB = ΔB * (∂B / ∂B)
= ΔB
ΔInet = ΔB * (∂B / ∂Inet)
= ΔB * (8μ0N) / √(125a)
ΔN = ΔB * (∂B / ∂N)
= ΔB * (8μ0Inet) / √(125a)
Δa = ΔB * (∂B / ∂a)
= ΔB * (-4μ0NInet) / (2√(125a^3))
These equations will give you the absolute errors for each variable in the equation B = (8μ0NInet) / √(125a), where Inet = I - Io.